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2 years ago

If December had 33 days, what would the 33rd calendar marker look like?

Mathematics

1 answer:

Maurinko [17]2 years ago

3 0

Answer:black

Step-by-step explanation:

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A car bought $52,000 and sold for $42,000. The percentage lost is?

mamaluj [8]

Calculate the difference between purchase and sales costs:

$52,000 - $42,000 = $10,000

Calculate what fraction of $52,000 is $10,000:

$\dfrac{\$10,000}{\$52,000}=\dfrac{10}{52}=\dfrac{5}{26}$

Transform it to the percent:

$\dfrac{5}{26}=\dfrac{5}{26}\cdot100\%=\dfrac{500}{26}\%\approx19\%$

<h3>Answer: A. 19%</h3>

8 0

2 years ago

Which table of values could be generated by the equation 10x + 5y = 15?

Ad libitum [116K]

Write the given equation AS FOLLOWS:
10x + 5y = 15
5y = 15 - 10x
Divide through by 5.
y = 3 - 2x

Test x=0: y = 3 - 2*0 = 3 => (0,3) Correct for B
Test x=1:. y = 3 - 2*1 = 1 => (1,1) Correct for B
Test x=2: y = 3 - 2*2 = -1 => (2,-1) Correct for B

Answer: B

3 0

2 years ago

Use green's theorem to compute the area inside the ellipse x252+y2172=1. use the fact that the area can be written as ∬ddxdy=12∫

Pavel [41]

The area of the ellipse $E$ is given by

$\displaystyle\iint_E\mathrm dA=\iint_E\mathrm dx\,\mathrm dy$

To use Green's theorem, which says

$\displaystyle\int_{\partial E}L\,\mathrm dx+M\,\mathrm dy=\iint_E\left(\frac{\partial M}{\partial x}-\frac{\partial L}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

( $\partial E$ denotes the boundary of $E$ ), we want to find $M(x,y)$ and $L(x,y)$ such that

$\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

and then we would simply compute the line integral. As the hint suggests, we can pick

$\begin{cases}M(x,y)=\dfrac x2\\\\L(x,y)=-\dfrac y2\end{cases}\implies\begin{cases}\dfrac{\partial M}{\partial x}=\dfrac12\\\\\dfrac{\partial L}{\partial y}=-\dfrac12\end{cases}\implies\dfrac{\partial M}{\partial x}-\dfrac{\partial L}{\partial y}=1$

The line integral is then

$\displaystyle\frac12\int_{\partial E}-y\,\mathrm dx+x\,\mathrm dy$

We parameterize the boundary by

$\begin{cases}x(t)=5\cos t\\y(t)=17\sin t\end{cases}$

with $0\le t\le2\pi$ . Then the integral is

$\displaystyle\frac12\int_0^{2\pi}(-17\sin t(-5\sin t)+5\cos t(17\cos t))\,\mathrm dt$

$=\displaystyle\frac{85}2\int_0^{2\pi}\sin^2t+\cos^2t\,\mathrm dt=\frac{85}2\int_0^{2\pi}\mathrm dt=85\pi$

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Notice that $x^{2/3}+y^{2/3}=4^{2/3}$ kind of resembles the equation for a circle with radius 4, $x^2+y^2=4^2$ . We can change coordinates to what you might call "pseudo-polar":